Question: $\dfrac{ n + 6p }{ -7 } = \dfrac{ -8n - 3q }{ -2 }$ Solve for $n$.
Answer: Multiply both sides by the left denominator. $\dfrac{ n + 6p }{ -{7} } = \dfrac{ -8n - 3q }{ -2 }$ $-{7} \cdot \dfrac{ n + 6p }{ -{7} } = -{7} \cdot \dfrac{ -8n - 3q }{ -2 }$ $n + 6p = -{7} \cdot \dfrac { -8n - 3q }{ -2 }$ Multiply both sides by the right denominator. $n + 6p = -7 \cdot \dfrac{ -8n - 3q }{ -{2} }$ $-{2} \cdot \left( n + 6p \right) = -{2} \cdot -7 \cdot \dfrac{ -8n - 3q }{ -{2} }$ $-{2} \cdot \left( n + 6p \right) = -7 \cdot \left( -8n - 3q \right)$ Distribute both sides $-{2} \cdot \left( n + 6p \right) = -{7} \cdot \left( -8n - 3q \right)$ $-{2}n - {12}p = {56}n + {21}q$ Combine $n$ terms on the left. $-{2n} - 12p = {56n} + 21q$ $-{58n} - 12p = 21q$ Move the $p$ term to the right. $-58n - {12p} = 21q$ $-58n = 21q + {12p}$ Isolate $n$ by dividing both sides by its coefficient. $-{58}n = 21q + 12p$ $n = \dfrac{ 21q + 12p }{ -{58} }$ Swap signs so the denominator isn't negative. $n = \dfrac{ -{21}q - {12}p }{ {58} }$